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there are $320 available to fence in a rectangular garden. the fencing for the side of the garden facing the road costs $15 per foot, and the fencing for the other 3 sides costs $5 per foot. the picture on the right depicts this situation. consider the problem of finding the dimensions of the largest possible garden.
a) determine the objective and constraint equations.
b) express the quantity to be maximized as a function of x.
c) find the optimal values of x and y

Respuesta :

Answer:

(a) The objective equation is [tex]A=xy[/tex] and the constrain equation is [tex]320=20x+10y[/tex]

(b) [tex]A=-2x^2+32x[/tex]

(c) The optimal values of x = 8 and y = 16

Step-by-step explanation:

Let y be the width and x the length of the rectangular garden.

(a) Determine the objective equation and the constraint equation.

  • In this problem we want to maximize the area of the garden. The objective equation is the area of the rectangular garden

[tex]A=xy[/tex]

  • The constraint is the amount of money we have for the fence and the constrain equation is

From the information given:

Cost of fence parallel to the road = $15x

Cost of the 3 other sides = $5(2y+x)

[tex]320=15x+5(2y+x)\\320=20x+10y[/tex]

(b) Express the quantity to be maximized as a function of x

We use the constraint equation to solve for y

[tex]20x+10y=320\\20x+10y-20x=320-20x\\10y=320-20x\\y=2\left(-x+16\right)[/tex]

Substituting into the objective equation

[tex]A=x\cdot (2\left(-x+16\right))\\A=-2x^2+32x[/tex]

(c) Find the optimal values of x and y

We have to figure out where the function is increasing and decreasing. Differentiating,

[tex]\frac{dA}{dx} =\frac{d}{dx}(-2x^2+32x)\\\\\frac{dA}{dx} =-\frac{d}{dx}\left(2x^2\right)+\frac{d}{dx}\left(32x\right)\\\\\frac{dA}{dx} =-4x+32[/tex]

Next, we find the critical points of the derivative

[tex]-4x+32=0\\-4x=-32\\x=8[/tex]

we need to make sure that this value is the maximum using the second derivative test:

if [tex]f''(x_0)<0[/tex], then f has a local maximum at [tex]x_0[/tex]

[tex]\frac{d^2A}{dx^2} =\frac{d}{dx} (-4x+32)=-4[/tex]

[tex]A''(8) < 0[/tex]

so x = 8 is a local maximum.

To find y,

[tex]y=2\left(-x+16\right)\\y=2\left(-8+16\right)\\y=16[/tex]

Ver imagen franciscocruz28

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